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Journal of Pure and Applied Algebra 89 (1993) 329-337 329 North-Holland

On biregular rings and their duality

Shu-Hao Sun* School of Mathematics FO7, University of Sydney, Sydney, NSW 2006, Australia

Communicated by G.M. Kelly Received 20 March 1992 Revised 20 November 1992

Abstract

Sun, S.-H., On biregular rings and their duality. Journal of Pure and Applied Algebra 89 (1993) 329-337.

We first give a new characterization of biregular rings which extends several existing results. Then we establish a new duality between biregular rings and compact Hausdorff simple ringed spaces which is closely related to the Dauns-Hofmann-Pierce duality between biregular rings and Stone simple ringed spaces.

Introduction

Dauns and Hofmann showed in [2, Theorem I] that biregular rings (i.e., those for which each principal ideal is generated by a central idempotent) can be represented as the rings of global sections of a ringed space (X, O), where X is a Stone space and 0 is a sheaf of simple rings over X; and that the ring of global sections of any sheaf of simple rings over a Stone space is biregular. Moreover, they showed that both assignments are functorial; and hence gave a duality between these two categories. Pierce in [9] gave a different proof. On the other hand, by observing the fact that biregularity implies that all primes are maximal while the converse is not necessary true, some authors were interested in the following question:

biregular = (pm) + ? ,

where (pm) stands for the property that all primes are maximal.

Correspondence to: S.-H. Sun, School of Mathematics F07, University of Sydney, Sydney, NSW 2006, Australia. * The author gratefully acknowledges the support of the Australian Research Council.

0022-4049/93/$06.00 @ 1993 - Elsevier Science Publishers B.V. All rights reserved 330 S.-H. Sun

Obviously, the commutativity of the ring is a sufficient condition. Kaplansky first observed in his unpublished paper (see also [l]) that in the realm of semiprime rings the property that any binary product of finitely generated ideals is finitely generated is a possible answer (Kaplansky called them neo-commutative rings). Goodearl showed in [3, Corollary 8.141 that if R is a regular ring satisfying general comparability, then R is biregular iff all primes are maximal. Very recently, Belluce showed in [l, Theorem 71 that Kaplansky’s condition above can be weakened to the condition that, for any two finitely generated ideals K, and K2, there is a finitely generated ideal K G K, K2 such that the prime radical of K1K2 is the prime radical of K; such rings will be called qc-rings. Since a regular ring satisfying general comparability is a qc-ring (see [l, Theorem 6]), Belluce’s result also extends the result above of Goodearl. Belluce also considered another condition (see [l, Theorem 91): if R is a ring without non-zero nilpotents, then R is biregular iff R satisfies pm. In Section 1 of this paper, we shall give an answer of which all the above become special cases. In Section 2, we shall use this new characterization to establish the duality between the category of biregular rings and the category of compact simple ringed spaces (that is, a compact ringed space, in the sense of [7], of which each stalk is a simple ring). In fact, we shall prove that the underlying space of each compact simple ringed space is a Stone space. In this paper, we always assume that a ring has an identity but is not necessarily commutative; and an ideal means a 2-sided ideal. The prime radical of an ideal Z i.e., the intersection of all prime ideals containing I) will be denoted by %‘/I (or L). Th e set of all (2-sided) ideals of a ring R will be denoted by Id R. For a E R, (a) will denote the principal ideal generated by the element a. Spec R and MaxSpec R will stand for the prime spectrum and the maximal spectrum of R (with the Zariski topology), respectively.

1. New characterizations

It is known that an ideal Z of a ring R is semiprime (i.e., J* c Z+ J c Z) iff Z = d/r (see, e.g., [4, Theorem 2.7]), and that the set SId R consisting of all semiprime ideals of R forms a complete Heyting algebra (in fact, it is isomorphic to the Zeriski topology of the prime spectrum Spec R of R). We would like to introduce the following.

Definition 1.0. For any ideal Z of a ring R, let I’ = c {J ) JZ C v/o}. An ideal Z is called tight if a E Z implies (u) ’ gZ. A ring R is called tight if each minimal prime ideal of R is tight.

We shall show that the following classes of rings are tight: (1) All qc-rings; in particular, all commutative rings, all biregular rings and all (right) Noetherian rings (see Corollary 1.9). On biregular rings and their duality 331

(2) All weakly symmetric rings in the sense of [lo]; in particular, all rings without non-zero nilpotents (see Proposition 1.11). (3) All prime rings (i.e., the zero ideal is prime); in particular, all free rings. Our main result in this section is that a ring R is biregular iff R is tight and semiprime with all primes maximal; which extends all results mentioned in the Introduction. For this purpose, we need some notions and notations. For a ring R, we write L(R) for the sub-meet lattice of Id R, generated by -\/r with I finitely generated. Then the typical element in L(R) is of the form gm, where Ij are finitely generated ideals of R. It is clear that L(R) is a lattice-in fact a distributive one-with qm for the join of I, and I,. Let R* denote the set of all such ideals having the form 1112 . . . I,, where each I, is finitely generated; then L(R) = {q/K 1K E R*}. We proved in [S, Proposition 1.71 the following:

Lemma 1.1. The following are equivalent for an ideal I of a semiprime ring R. (a) I is complemented in L(R). (b) I is complemented in Id R. (c) I is a principal ideal generated by a central idempotent. Cl

Following Arens and Kaplansky, a ring R is called biregular if each principal ideal is generated by a central idempotent. Clearly, in a biregular ring R, each finitely generated ideal is too generated by a central idempotent (e.g., e,R + e,R = (e, + e2 - e,e,)R); and hence each finite product of finitely generated ideals is generated by a central idempotent. Thus we have the following:

Lemma 1.2. A ring R is biregular if and only if each ideal in R* is generated by a central idempotent if and only if R* = L(R) is the Boolean algebra consisting of all principal ideals of R. q

Using Lemma 1.1 and Lemma 1.2, we further have the following:

Theorem 1.3. A ring R is biregular if and only if R is semiprime and L(R) is a Boolean algebra. 0

For each proper ideal I of L(R), define 4(I) = v/c { K / K E I}. Then +(I) is a proper semiprime ideal of R and further we have the following lemma:

Lemma 1.4. For any ring R, the map 4 is a frame morphism from Id L(R) to SId R (i.e., a map preserving finite meets and arbitrary joins).

Proof. Let I and J be ideals of L(R). Then 332 S.-H. Sun

w fl 4 c W) f-l 4(J) =Q{KIKEZ}C{LILEJ}

c C{iv(NEzf-u}=+(zn.q.

For each v/K E V Zi, we have

where each v/K, E Ii.; and hence I

d/K = j/dK, v . . .4/K, c V +(Zi,) since each q/K, c 4(Z,,). This completes the proof. 0

Thus C#Jhas a right adjoint #J* which is determined by

4*(K) = V {ZE Id L(R) I 440 c W > for all K E SId R. Since C#Jpreserves finite meets, +* preserves prime elements; i.e., the restriction of 4*, denoted by m, is a mapping from Spec R to Spec L(R). More precisely, m is defined by m(P) = {I E L(R) 1Z C P} for each P E Spec R.

Lemma 1.5. (1) 4*(K) = {J E L(R) 1J c K} fur each K E SId R. (2) The map m : Spec R+ Spec L(R) is an embedding.

Proof. (1) is easy to check. (2) follows from the fact that +* is a right adjoint to 4 and that 4 is onto and Spec L is T,; or more directly, it follows from the fact that P E Zl((a)) of Spec R iff +*(P) E D(m) of Spec L(R) for all a E R, where D(Z) is a typical open set of Spec R (or Spec L(R)). (For the fact that m is injective see also [ 11.) q

Belluce showed in [l, Theorem l] that a ring R is qc iff m is a homeomorphism. However, we have the following:

Theorem 1.6. For any ring R, the restriction of m gives a homeomorphism between MaxSpec R and MaxSpec L(R).

Proof. Let M be a maximal ideal of R. Then m(M) = +*M is a proper ideal of L(R). If it is not maximal, then there exists a maximal ideal N of L(R) which properly contains m(M). Thus 4(N) is a proper ideal which contains M; therefore #(N) = M and &*(#(A’)) = m(M). But 4*(4(N)) > N-this is a contradiction. Thus we have shown that the restriction of m is well-defined. It remains to show On biregular rings and their duality 333 that m is surjective. Let N be a maximal ideal of L(R). Then 4(N) is a proper ideal of R and hence there esists a maximal ideal A4 > 6(N). Thus m(M) > m(+(N)) 2 N and hence m(M) = N since N is maximal. This completes the proof. 0

On the other hand, the map m does not necessarily preserve minimal prime ideals in general. We will show below that R is tight if and only if m preserves minimal primes. Similarly to Definition 1.0, an ideal Z of a distributive lattice L is called tight if for each a E Z there is a b j5! P with b A a = 0. Note that, for any K,Z,JEI~ R, K = Z+ J implies that K’ = I’ f~ J’. So we have the following:

Lemma 1.7. Let P be a prime ideal of a ring R. Then P is tight iff m(P) is tight.

Proof. Let P be tight. If m(P) contains v’K and the set {d.Z E L(R) 1l/J fl I,’ K = v’O} for some K E R*, then %‘K C P by the definition. Moreover, for each XER with (~).k’\/K~fi, we have mnqK=vO and mEm(P); and hence x E P; that is, K’ C P. Let K = K, K2 . . . K,, where each Ki is a finitely generated ideal of R. Then there is Ki C P since P is prime and hence KI C K’ r P. Furthermore, let K, = (a,) + (a,) + *. . + (a,). Then KI = (a,)’ fl (a,)‘n... n (a,,)’ c P and hence there is a (ai)’ c P, for some Z, since P is prime. On the other hand, we also have ( aj) c Kj c P, which means that P is not tight-a contradiction. For the converse, suppose that P contains both a and (a)l, then mu m(P). For each d/K E L(R) with %‘K fl m = -\/O, we have K + (a) c d/o and hence KC (a) 1 C P, v/K E m(P)-which means that m(P) is not tight. This completes the proof. 0

Let MinSpec R (resp. MinSpec L(R)) denote the subspace, consisting of all minimal prime ideals, of Spec R (resp. Spec L(R)).

Observation. Let P be a prime ideal of a distributive lattice L. Then P is minimal if and only if it is tight.

Proof. It is clear that a tight prime ideal is minimal. Now suppose that P is a minimal prime ideal and there is a a, E P with a, A b # 0 for any b e P. Then the set (L\P) U {a,} can be generated a maximal filter, say M, of L and hence L\M is a prime ideal which is properly contained in P-a contradiction. q

Thus, we have the following proposition:

Proposition 1.8. For any ring R, the following are equivalent: 334 S.-H. Sun

(1) m(MinSpec R) c MinSpec L(R). (2) R is tight.

Proof. Suppose (1). Then for each minimal prime ideal P of R, m(P) is a minimal prime of L(R) and hence is tight. It follows from Lemma 1.7 that P is tight. Conversely, if R is tight, that is, each minimal prime ideal of R is tight, then m(P) is a tight prime ideal of L(R) by Lemma 1.7, and hence is minimal. •i

If R is qc, then, by [l, Theorem 11, m is a homeomorphism, in particular, R satisfies the condition (1) in Proposition 1.8. Thus we have the following corollary:

Corollary 1.9. Each qc-ring R is tight. 0

The converse is not true (see the Remark of Proposition 1.11). Now we can prove our main result.

Theorem 1.10. For a ring R, the following are equivalent: (a) R is biregular. (b) R is tight and semiprime with all primes maximal.

Proof. We need only show that (b) implies (a). By Theorem 1.3, it suffices to show that L(R) is Boolean; or equivalently, that all primes of L(R) are maximal. Let P be a prime ideal of L(R). Then there is a maximal ideal M of R such that m(M) 2 P by Theorem 1.6. But by the assumption, A4 is also a minimal prime ideal of R; so that m(M) is a minimal prime ideal of L(R) by Proposition 1.8. Thus P = m(M) is maximal (is also a minimal prime ideal) and hence L(R) is Boolean. q

We shall end this section by showing that all weakly symmetric rings, in particular, all rings without non-zero nilpotents, are tight. This result allows us to easily give examples which is tight but not qc. Following Lambek [6], an ideal I of a ring R is called symmetric if abc E Z j acb E Z and R is called symmetric if the zero ideal is symmetric. Recall from [lo] that R is called weakly symmetric if the prime radical V’O contains a symmetric ideal. Clearly, the class of symmetric rings is contained in that of weakly symmetric rings, and contains all rings without non-zero nilpotents. Various examples show that a weakly symmetric ring need not be symmetric, while a symmetric ring may have non-zero nilpotents (see [6] and [lo]).

Proposition 1.11. Zf R is a weakly symmetric ring, then each minimal prime ideal of R is tight. On biregular rings and their duality 335

Proof. First suppose that R has no non-zero nilpotents. Then, for each a E R, Ann a = Ann(u). Note that in this case I’ = Ann I. Let P be a prime ideal containing a and Ann a. Let Q = R\P and let M be the multiplicative set generated by Q and a. Then M does not contain 0. Let Y be the set of multiplicative sets which contain M but do not contain 0. Using Zorn’s Lemma, there exists a maximal element M, in Y, and hence there exists a prime ideal P, which is properly contained in P. That is, P would not be a minimal prime. Now suppose that R is weakly symmetric. Since R/v0 is a ring without non-zero nilpotents, each minimal prime ideal of R/d0 does not contain both a E R/k’0 and Ann(a) in RlvO. It easily follows that each minimal prime ideal of R is also tight. q

Remark. Not all weakly symmetric rings are qc. For example, let R be the free ring on two generators. Then R is not qc (for details see [ll]). Note that a free ring is also trivially a prime ring while a prime ring is trivially tight.

2. Duality

In this section, we shall use Theorem 1.10 to prove that the category of compact Hausdorff simple ringed spaces is isomorphic to the category of Stone simple ringed spaces; and hence is dual to the category of biregular rings and their morphisms . Recall that a ringed space (X, S) is called Stone simple if X is a Stone space and each stalk of 9 is a simple ring.

Definition 2.1. A ringed space (X, 9) is called Huusdorff if for any distinct x1 and x2 of X there is a global section p such that p(xl) = 0 and p(x2) = 1; if in addition, X is a compact space, then (X, 9) is called a compact Hausdorff ringed space. Following Mulvey [7], a ringed space (X, 9) is called completely regular if for any x E X and any closed subset F of X with xg F there is a global section p, which is the identity at x and vanishes on F.

Lemma 2.2. Each compact Huusdorff ringed space is completely regular.

Proof. Let x$ZF, where F is a closed subset. Then for each y E F there is a section pr such that p,(x) = 1 and p,(y) = 0. Thus there is an open neighborhood V, of y such that pYvanishes on V, . Since {V,, 1 y E F} is an open cover of F there is a finite subcover {VYi 1i 5 n}. Now let p = pY,pYz . . * py,. Then p is the identity at x and vanishes on F; as required. 0

In [9], Pierce showed that for each Stone ringed space the ring of global sections can separate points and closed sets; hence it is a compact Hausdorff 336 S.-H. Sun ringed space. But as expected, the converse is not true. For example, given a compact Hausdorff space which is not a Stone space, say the closed interval [0,11 of the real line, consider the sheaf 9 determined by all continuous real value functions on it. Then ([0, 11, 9) is a compact Hausdorff ringed space while it is not Stone. However, with a little surprise, we shall prove that a compact Hausdorff simple ringed space is precisely a Stone simple ringed space; and hence give a duality between the category of biregular rings and the category of compact Hausdorff simple ringed spaces. Let (X, 9) be a compact Hausdorff simple ringed space. We shall prove that the ring A of global sections of %is biregular and that the prime spectrum Spec A of A is homeomorphic to X. Thus X is actually a Stone space. We need some lemmas. For each x E X, let Z, = {p E A ( p(x) = O}. Then Z, is an ideal of A. (For the following lemma see also [7].)

Lemma 2.3. For each prime P of A there is an x E X such that P 2 Z,.

Proof. If not, for each x E X there exists a section p, E A such that p,(x) = 0 and p, g P. Hence there is an open neighbourhood V, of x on which p, vanishes. By compactness, there is finite cover {Vxi ) i I n} of X. It is not hard to see that

P,,AP,~A. . . AP,~ = 0 2 which contradicts the fact that all the p,$?P and that P is prime. 0

Lemma 2.4 [7, Theorem 1.41. The stalk Sx of 9 at x is isomorphic to A/Z,. q

Now by using Theorem 1.10 and some lemmas in this section, we have the following:

Theorem 2.5. Let A be the ring of global sections of a compact Hausdorff simple ringed space (X, 9). Then A is biregular and Spec A is homeomorphic to X.

Proof. Let P be a prime ideal of A. Then, by Lemma 2.3, P > Z, for some x E X and hence P/Z, is a proper ideal of 5x by Lemma 2.4. Thus P/Z, is O,, or equivalently, P = Z,. This also proves that each prime is maximal. Moreover, for all x E X, Z, is prime and so n {P E Spec A} = 0, the zero section; and hence A is semiprime. To show that A is biregular, it suffices, by Theorem 1.10, to show that each Z, is tight. Let p E Z, ; then p(x) = 0 and hence there is an open neighborhood V of x on which p vanishes. By Lemma 2.2, there is a section 4 E A such that 4(x) = 1 and vanishes on the complement of V. Then 4 eZ, and +Ap = 0, so Z, is tight. Finally, sending each Z, to x gives a homeomorphism between Spec A and X. WI bwegular rrngs and their duality 337

Now the conclusion follows from the fact that Spec A is a Stone space. This completes the proof. 0

Daun and Hofmann, and Pierce showed that the category of biregular rings, and ring morphisms preserving central idempotents, is dual to the category of Stone simple ringed spaces. So we have the following theorem:

Theorem 2.6. The category of biregular rings and their morphisms is dual to the category of compact Hausdorff simple ringed spaces.

Remark 2.7. By extending the ideas and techniques here, we will prove that the category of tight weak Baer rings and their morphisms is dual to the category of compact Hausdorff prime ringed spaces (for details see the author’s [12]), where a ring is called weak Baer if for each a E R the left annihilator Ann,(a) is generated by a central idempotent.

Acknowledgment

The author wishes to thank Professor Max Kelly for his many valuable suggestions and encouragement. He would also like to thank the referee for his many valuable comments which improved this paper.

References

[l] L.P. Belluce, Spectral spaces and non-commutative rings, Comm. Algebra 19 (1991) 1855-1865. [2] J. Dauns and K.H. Hofmann, The representation of biregular rings by sheaves, Math. Z. 91 (1966) 103-123. [3] K.R. Goodearl, Von Neuman Regular Rings (Pitman, London, 1979). [4] K.R. Goodearl and R.B. Warfield, Jr, An Introduction to Noncommutative Noetherian Rings, London Mathematical Society Student Texts, Vol. 16 (Cambridge University Press, Cambridge, 1989). [5] I. Kaplansky, Topics in commutative ring theory, Dept. of Mathematics, University of Chicago. [6] J. Lambek, On the representation of modules by Sheaves of factor modules, Canad. Math. Bull. 14 (1971) 459-466. [7] C.J. Mulvey, Compact ringed spaces, J. Algebra 52 (1978) 411-436. [S] S.B. Niefield and S.-H. Sun, Algebraic De Morgan’s laws for noncommutative rings. Submitted for publication. [9] R.S. Pierce, Modules over Commutative Regular Rings, Memoirs of the American Mathematical Society, Vol. 70 (American Mathematical Society, Providence, RI, 1967). [lo] S.-H. Sun, Non-commutative rings in which each prime ideal is contained in a unique maximal ideal, J. Pure Appl. Algebra 76 (2) (1991) 179-192. [ll] S.-H. Sun, On the least multiplicative nucleus of a ring, J. Pure Appl. Algebra 78 (3) (1992) 311-318. [12] S.-H. Sun, Duality on compact prime ringed spaces, Submitted for publication.